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butalik [34]
3 years ago
14

What is x X/5=15. ⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️⌚️

Mathematics
2 answers:
Debora [2.8K]3 years ago
7 0

Answer:

x=75

Step-by-step explanation:

\frac{x}{5} =15

x=15 \times 5

x=75

kotegsom [21]3 years ago
4 0

Answer:

x=3

Step-by-step explanation:

15/5=3 now I gotta get 20 chatacters

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g Given the following premises: (1) a∧(b∨a) (2) ~c→~a (3) ~g→~a (4) g→e (5)~(d∨h) Prove the Conclusion: c∧~h?
KIM [24]

Answer:

See the argument below

Step-by-step explanation:

I will give the argument in symbolic form, using rules of inference.

First, let's conclude c.

(1)⇒a  by simplification of conjunction

a⇒¬(¬a) by double negation

¬(¬a)∧(2)⇒¬(¬c) by Modus tollens

¬(¬c)⇒c by double negation

Now, the premise (5) is equivalent to ¬d∧¬h which is one of De Morgan's laws. From simplification, we conclude ¬h. We also concluded c before, then by adjunction, we conclude c∧¬h.

An alternative approach to De Morgan's law is the following:

By contradiction proof, assume h is true.

h⇒d∨h by addition

(5)∧(d∨h)⇒¬(d∨h)∧(d∨h), a contradiction. Hence we conclude ¬h.    

8 0
3 years ago
Can you help me pleaseeeeeew
Lera25 [3.4K]

Answer:

<h2>The answer is A 14.6</h2>

Step-by-step explanation:

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4 0
3 years ago
Whats the square root of 826 also looking for a minecraft girI needs to be 15 or older
skad [1K]

Answer:

28.740,

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
2 years ago
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The length of the trail is 98.5 KM, all you need to do is divide the total distance by the number of times she went on it.
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3 years ago
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